voracious | video player for studying foreign languages | Frontend Framework library
kandi X-RAY | voracious Summary
kandi X-RAY | voracious Summary
Voracious is mostly built as a single-page web app with React (using create-react-app), and then packaged as a cross-platform desktop app using Electron. As with a normal create-react-app-based app, the bulk of the code is in src/ with some static resources in public/. The output of webpack (after yarn react-build) will go into build/. Electron-builder is used to package the resulting build into an Electron-based app, with its output going to dist/. The app/ dir contains a small amount of code and config related to Electron. The main Electron entry point is app/electron-main.js, and app/package.json is the package.json that's distributed in the final app. Most third-party dependencies are pure JS (react, immutable, etc.) and are declared in the root package.json. Those are bundled with the the Voracious code (by webpack) into a single JS file. That means that the root node_modules/ does not needed to be distributed with the final Electron app. This sort of bundling isn't generally necessary to do for Electron apps, but it's convenient for various reasons. Dependencies that use native code (e.g. sqlite) need to be compiled against the Electron V8 runtime, and are declared in app/package.json. The corresponding app/node_modules/ is packaged into the final distributed Electron app. Electron-builder is configured via electron-builder.json. The current config has it basically combine the contents of build/ and app/ to form the distributed archive.
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Top functions reviewed by kandi - BETA
- Queue task queue .
- Iterate over an object
- Registers handlers for the IPC process .
- Calls a function with an array of promises
- Insert data into task
- Dictionaries container for Tokenizer
- Removes path from root to an array .
- Creates a new Browser window
- Create a test function
- loop for iteration
voracious Key Features
voracious Examples and Code Snippets
Community Discussions
Trending Discussions on voracious
QUESTION
I'm trying to enrich a web site with Open Graph data, to let the page author decide which image and texts show up in a preview on social media. However, I can't get WhatsApp to show a small image in the sharing card.
My current markup looks like this:
...ANSWER
Answered 2019-Sep-14 at 17:24Your og implementation only works on some versions of whats app. And that is because the url (of the entire page, not just the image) contains non ascii characters.
I have copy/paste your html code on one of my web-sites and it worked ok, as long as the file was named index.html or ogtest.html.
After I renamed the file with Greek characters, og-protocol stopped working.
QUESTION
I'm attempting Codility fish challange which is described as follows:
You are given two non-empty arrays A and B consisting of N integers. Arrays A and B represent N voracious fish in a river, ordered downstream along the flow of the river.
The fish are numbered from 0 to N − 1. If P and Q are two fish and P < Q, then fish P is initially upstream of fish Q. Initially, each fish has a unique position.
Fish number P is represented by A[P] and B[P]. Array A contains the sizes of the fish. All its elements are unique. Array B contains the directions of the fish. It contains only 0s and/or 1s, where:
0 represents a fish flowing upstream, 1 represents a fish flowing downstream. If two fish move in opposite directions and there are no other (living) fish between them, they will eventually meet each other. Then only one fish can stay alive − the larger fish eats the smaller one. More precisely, we say that two fish P and Q meet each other when P < Q, B[P] = 1 and B[Q] = 0, and there are no living fish between them. After they meet:
If A[P] > A[Q] then P eats Q, and P will still be flowing downstream, If A[Q] > A[P] then Q eats P, and Q will still be flowing upstream. We assume that all the fish are flowing at the same speed. That is, fish moving in the same direction never meet. The goal is to calculate the number of fish that will stay alive.
For example, consider arrays A and B such that:
A[0] = 4 B[0] = 0 A1 = 3 B1 = 1 A2 = 2 B2 = 0
A[3] = 1 B[3] = 0 A[4] = 5 B[4] = 0 Initially all the fish are alive and all except fish number 1 are moving upstream. Fish number 1 meets fish number 2 and eats it, then it meets fish number 3 and eats it too. Finally, it meets fish number 4 and is eaten by it. The remaining two fish, number 0 and 4, never meet and therefore stay alive.Write a function:
int solution(vector &A, vector &B);
that, given two non-empty arrays A and B consisting of N integers, returns the number of fish that will stay alive.
For example, given the arrays shown above, the function should return 2, as explained above.
Write an efficient algorithm for the following assumptions:
N is an integer within the range [1..100,000]; each element of array A is an integer within the range [0..1,000,000,000]; each element of array B is an integer that can have one of the following values: 0, 1; the elements of A are all distinct.
My solution is as follows:
...ANSWER
Answered 2019-Sep-18 at 18:32I'm going to post some algorithm thoughts and ideas for you. Hopefully this will help you visual the problem.
Some initial observations - all fish swimming upstream at the beginning of the array can never eat or be eaten. Same for fish at the bottom of the array swimming downstream. The action occur when scanning from front to back when a downstream meets an up.
First, walk your array front to back until you find a down swimming fish.
Next continue walking your down/up array until it swaps from down to up.
Resolve combat and increment a combat counter.
If down wins, keep walking down. If up wins, save the index of your location in the array(s), then walk it up the array until it is eaten or you hit a fish swimming the same direction.
If it finds a fish swimming its direction, it will live, jump back to where it started scan downwards again.
If it is eaten, take this fish and start comparing where you left off in the down array.
Repeat until you reach the end of the list. Return the (starting number of fish - number of combats)
The neat part is this is in-place. No moving of memory, no need to mark dead fish, etc. It will run linear time - at most nearly 2x linear.
This the the pseudo code I would use to resolve this.
QUESTION
I was working on a Codility problem:
...You are given two non-empty zero-indexed arrays A and B consisting of N integers. Arrays A and B represent N voracious fish in a river, ordered downstream along the flow of the river.
The fish are numbered from 0 to N − 1. If P and Q are two fish and P < Q, then fish P is initially upstream of fish Q. Initially, each fish has a unique position.
Fish number P is represented by A[P] and B[P]. Array A contains the sizes of the fish. All its elements are unique. Array B contains the directions of the fish. It contains only 0s and/or 1s, where:
0 represents a fish flowing upstream, 1 represents a fish flowing downstream. If two fish move in opposite directions and there are no other (living) fish between them, they will eventually meet each other. Then only one fish can stay alive − the larger fish eats the smaller one. More precisely, we say that two fish P and Q meet each other when P < Q, B[P] = 1 and B[Q] = 0, and there are no living fish between them. After they meet:
If A[P] > A[Q] then P eats Q, and P will still be flowing downstream, If A[Q] > A[P] then Q eats P, and Q will still be flowing upstream. We assume that all the fish are flowing at the same speed. That is, fish moving in the same direction never meet. The goal is to calculate the number of fish that will stay alive.
ANSWER
Answered 2018-Mar-14 at 23:57It's hard to follow this code with the strange data structures and lack of variable names, but I think I have the needed understanding ...
Space complexity:
The only dimensioned space you have is myQ, which is bounded by the total quantity of fish. Thus, this is O(n).
Time Complexity:
Given your strange logic, this was harder to follow. The paired decrements of remFish and the abuse of while -- break confused me for a couple of minutes. However, the simpler analysis is ...
The while -- break turns that loop into a simple if statement, since you always break the loop on first iteration. Therefore, your only true iteration is the for loop, bounded by the quantity of fish. Thus, this is also O(n).
Among other properties, note that you decrement numFish on each iteration, and it never drops as far as 0.
Why do you gauge one iteration on a.length and another on b.length? Those must be the same, the starting quantity of fish.
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